Example

Example: Graphing 4x29y224x36y36=04x^2 - 9y^2 - 24x - 36y - 36 = 0

To graph the hyperbola 4x29y224x36y36=04x^2 - 9y^2 - 24x - 36y - 36 = 0, first write the equation in standard form by completing the squares. Group the variables to obtain 4(x26x)9(y2+4y)=364(x^2 - 6x) - 9(y^2 + 4y) = 36. Complete the square inside the parentheses by adding 99 to the xx-terms and 44 to the yy-terms, balancing the right side: 4(x26x+9)9(y2+4y+4)=36+36364(x^2 - 6x + 9) - 9(y^2 + 4y + 4) = 36 + 36 - 36. This yields 4(x3)29(y+2)2=364(x - 3)^2 - 9(y + 2)^2 = 36. Divide each term by 3636 to obtain the constant 11, resulting in the standard form (x3)29(y+2)24=1\frac{(x - 3)^2}{9} - \frac{(y + 2)^2}{4} = 1. Because the x2x^2-term is positive, the hyperbola opens left and right. The center (h,k)(h, k) is (3,2)(3, -2), with constants a=3a = 3 and b=2b = 2. Sketch the central rectangle by moving 33 units left and right of the center, and 22 units above and below. Draw the asymptotes through the rectangle's diagonals, mark the vertices at (0,2)(0, -2) and (6,2)(6, -2), and graph the branches passing through these vertices.

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Updated 2026-05-25

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